A mixture of 010 mol of NOg 005 mol of H2g and 010 mol of H2

A mixture of 0.10 mol of NO(g), 0.05 mol of H2(g), and 0.10 mol of H2O(g) is placed in a 1.0-L vessel at 300K?

The following equilibrium is established: 2NO(g) + 2H2(g) <---> N2(g) + H2O(g). At equilibrium [NO] = 0.062M. Calculate Kc, and Kp.

Solution

2 NO + 2 H2 ==> N2 + 2 H2O

First I would set up an ICE box, which is a table showing the Initial concentrations, the Change in those concentrations, and the Equilibrium concentrations.

In the first row of the ICE box, I would put 0.10 mol NO, and 0.05 mol H2 for the reactants, and then put 0 mol N2 and 0.10 mol H2O for the products.

If at equilibrium, the NO becomes 0.062 mol, then it means that the amount of NO molecules DECREASED by 0.10-0.062 = 0.038.

The other chemicals need to change like this:

H2 decreases by 0.038 so it now is 0.05 - 0.038 = 0.012 M

N2 increases by HALF as much as the other guys because its coefficient is only 1. So N2 increases from 0 to 0.019 M

H2O increases by 0.038, so it is now 0.1 + 0.038 = 0.138 M